Structural analysis prep

cgcsb

Ok, so basically I need someone to tell me where I'm going wrong and how I can improve. So attached is a copy of the exam paper, below that is my attepmt, pitiful it may be. I can't access the library in Bolton st because it's being renovated 2 weeks before the exams :mad:

So I figure for Q1 a) the vertical reactions at joint a and b are 50kn and there is a horizontal reaction at joint b of 20kn.

b)so to determine the force in member a I said that:

the sum of the forces in the Y direction= 50-Acos30, and from that the force in A=57.735kn.

applying the same logic to K, I got K=57.735kn

The problem here is B and J (don't laugh). for B, as far as I can see has an internal force of 0 because there is no horizontal load and a horizontal member cannot sustain a vertical force.

As for J It's only horizontal force is 20kn (the reaction force as a result of the 20kn in the top right corner) so the internal force is 20kn??

c) SO using the method of sections, I cut through members D, E and F and disregarded everything to the right of the cut. So I am only paying attention to the left side of the frame.

The external forces acting on the left side of the frame are 50kn upwards and 40kn downwards. There are no horizontal forces. So knowing this I can say:

The sum of the forces in the y direction are: 50-40 +E(sin60)=> E= -11.547, the minus indicates that the force is acting downwards in this case.

I have no idea how to go about finding the forces in D and F

d) I'm willing to bet that A, D and K are Struts where as E, F and J are ties.

Q2
a) For this question Iconverted all the UDLs to point loads.

so from left to right: 6kn/m*2m gives us a point load of 12kn @ 1m to the left of point B (the centre of gravity of section AB)

14kn/m*7.2m gives us a point load of 100.8kn @ 3.6m to the right of point B

we also have 18kn point load @1.2m to the right of point C

all together we have a downward force of 130.8kn. So presumably, the opposite reaction is the same i.e. Rb+Rc=130.8kn

so taking the moments about Rb=0=-12(1)+100.8(3.6)-Rc(6)+18(7.2)
Rc=80.08kn
Rb+80.08=130.8 => Rb=50.72kn

I can draw the diagrams for Q2 without any hastle and I'm ok with them. So I'm not gonna bother skanning in my diagrams for them.

Q3 for this type of question, I was actually really good at them about a year ago but the oul memory isnt the best can someone give me a hand with Q3, thanks in advance guys

owenmul

hi, i'll do this out for you tomorrow, shouldn't take me long, i can email it to you or i can post it here, whichever suits

cgcsb

thankyou very much